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Consider a real-valued function $f$ on $[a,b]$ with the property that for every $\epsilon>0$, the set $\{x \in[a,b] : |f(x)| \geq \varepsilon\}$ is finite. So, in other words, $f$ is zero almost everywhere on its domain. It seems intuitively true that $$ \int_{a}^{b} f \, dx =0, $$ but I'm having a hard time proving even that $f \in \mathscr{R}([a,b])$.

So far I got: Let $\varepsilon>0$. I want to show that there is a partition $P$ such that $U(P,f)-L(P,f)<\varepsilon$. Choose the partition $P=\{ x_{i} : i =0,1,\dots,n \}$ containing the points where $|f(x_{i})|\geq \varepsilon$. Then $$ L(P,f) = \sum_{ i=1 }^{ n } m_{i} \Delta x_{i} < \varepsilon (b-a). $$

I don't know how to proceed here. Any help would be greatly appreciated!

Incubu121
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  • @geetha290krm Can you help me understand how the question you linked is relevant to the statement I want to prove? – Incubu121 Aug 17 '24 at 09:54
  • @geetha290krm I'm afraid the Wikipedia page for the Riemann integral is a little bit too broad to be of any help. Can you direct me to the section of the page you have in mind? – Incubu121 Aug 17 '24 at 09:57
  • @geetha290krm According to Wikipedia, "Riemann's Theorem" is considered to refer to the theorems mentioned in https://en.wikipedia.org/wiki/Riemann%27s_Theorem. None of these seems to relate to my question. Searching for "Riemann's theorem for integrals" and similar queries does not provide an answer to my question. – Incubu121 Aug 17 '24 at 11:50
  • @geetha290krm No, that is not a counter-example. Fix $\varepsilon=\frac{1}{2}$. Then, for all $a<b$, the set $\left{ x \in [a,b] : |f(x)| \geq \varepsilon \right}$ is infinite, not finite. This is because the rational numbers are dense in $\mathbb{R}$. – Incubu121 Aug 17 '24 at 12:13
  • I guess what got geetha290krm is that almost everywhere has a technical meaning that allows infinite sets. Including their example. – Jyrki Lahtonen Aug 17 '24 at 12:18
  • Search gives this. A special case. – Jyrki Lahtonen Aug 17 '24 at 12:22
  • @geetha290krm I apologize for using a technical term out of place in the title, but obviously this is not what I'm asking? You have written multiple comments so far (most of which you have deleted), none of which have been helpful in answering the actual question. I appreciate your time, but it might be better to stop now. – Incubu121 Aug 17 '24 at 13:11

2 Answers2

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The general idea is easy: For a given $\varepsilon>0$, you need to do two things:

  • Make sure the intervals around the "problematic points" are narrow enough that $U-L$ is at most $\varepsilon/2$ in total for these intervals
  • Make sure the rest of is such that $U-L$ on this part is at most $\varepsilon/2$ in total for these intervals

That's the idea. So let's get to it.

First off, we decide what our problematic points will even be. Because of point 2, it is most immediate to choose $\{x:|f(x)|\geq\frac{\varepsilon}{4(b-a)}\}$. This is just enough to relatively easily ensure the $U-L$ bound we need on the unproblematic part.

Now to make sure that the intervals around the problematic points are narrow enough. Let $x_1, x_2,\ldots,x_n$ are the problematic points. Let the partition consist of $a, b$, and all points of the form $$ x_k\pm\frac{\varepsilon}{4n|f(x_k)|} $$ that lie between $a$ and $b$.

It is relatively straight-forward to calculate that this partition gives the $U-L$ we desire. A slight complication arises if one of the problematic points turns out to be exactly one of the partition points. I suggest making the $\pm$ term above slightly smaller to avoid this.

Arthur
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  • We could use $$x_k \pm \min!\left( \frac{\varepsilon}{2n|f(x_k)|}, \frac 13 \min_{i \neq k} |x_i-x_k|\right)$$ – aschepler Aug 17 '24 at 13:06
  • Thank you, that's very easy to understand! I got it to work with $\left{x:|f(x)| \geq \frac{\varepsilon}{2(b-a)}\right}$ and $x_k \pm \frac{\varepsilon}{4 n\left|f\left(x_k\right)\right|}$, might this be what you where going for? I posted a solution where I try to write the whole thing out, does it look correct to you? – Incubu121 Aug 17 '24 at 13:22
  • @Incubu121 We need 4 on both denominators, actually. We want $U-L\leq \varepsilon/2$ separately on both the problematic points and the rest of the interval, so want ye width in terms of $\varepsilon/2$. But here are working directly with the "radius" instead, so we need ${}/4$. – Arthur Aug 17 '24 at 16:42
  • @Arthur It really seems to me that we only need $/4$ when forming the intervals. If we assure that $|f(x)|<\frac{\varepsilon}{2(b-a)}$ for non-problematic points, then the contribution from the intervals that contain no problematic points are $$ \sum\left(M_i-m_i\right) \Delta x_i<\sum \frac{\varepsilon}{2(b-a)} \Delta x_i\leq\frac{\varepsilon}{2}. $$ What am I missing here? – Incubu121 Aug 18 '24 at 09:36
  • We only know that $$\frac{\varepsilon}{2(b-a)}> M_i\geq m_i> -\frac{\varepsilon}{2(b-a)}$$ It is very much possible for $M_i-m_i$ to be closer to $\frac{\varepsilon}{b-a}$. So we need another factor of $\frac12$. Maybe we even need $/8$ around the problematic points, for basically the same reason (the possibility that $x_i$ and $x_{i+1}$ are in the same interval and $f(x_i)$ and $f(x_{i+1})$ have opposite signs, or just that $|f(x_i)|$ is close to $\frac{\varepsilon}{2(b-a)}$). I haven't thought that one quite through. – Arthur Aug 18 '24 at 11:41
  • @Arthur But mustn't $m_i$ be exactly 0 for any conceivable subinterval? Otherwise, we would contradict the fact that for every $\varepsilon>0$, only a finite number of points in $[a, b]$ (and therefore in any subinterval thereof) are such that $|f(x)| \geq \varepsilon$. Thus $M_i-m_i = M_i$, and my argument should hold. – Incubu121 Aug 18 '24 at 12:11
  • No. $m_i$ is the smallest value of $f(x)$ on any given interval. It is not the smallest value of $|f(x)|$ or something. For instance, for the function $f(x)=x$, on the interval $[-1,1]$, we have $M_i=1$, $m_i=-1$, which means $M_i-m_i=2$. Thus if $|f(x)|$ is bounded by some $K$, then $|M_i|$ and $|m_i|$ are bounded by the same $K$, but $M_i-m_i$ is bounded by $2K$. – Arthur Aug 18 '24 at 12:39
  • For an example that better suits our particular requirements for $f$, we will of course have to make it discontinuous. But nothing in this setup stops negative function values. – Arthur Aug 18 '24 at 12:42
  • @Arthur Yep, I see it now. I appreciate your patience. – Incubu121 Aug 19 '24 at 08:54
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Let $$f_n(x)=\begin{cases} f(x)& |f(x)|\ge n^{-1} \\ 0 &|f(x)|<n^{-1} \end{cases}$$ The function $f_n$ is Riemann integrable as $f_n(x)=0$ except for finitely many $x.$ Moreover $f_n\rightrightarrows f,$ because $|f_n(x)-f(x)|\le n^{-1}.$ Hence $f$ is Riemann integrable. As $\int_a^bf_n(x)\,dx =0$ we get $\int_a^bf(x)\,dx=0.$

Ryszard Szwarc
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